a. \(x^2+6x+5=0\)
⇔ \(x^2+5x+x+5=0\)
⇔ \(\left(x^2+x\right)+\left(5x+5\right)=0\)
⇔x\(\left(x+1\right)+5\left(x+1\right)=0\)
⇔ \(\left(x+1\right)\left(x+5\right)=0\)
⇔\(\left[{}\begin{matrix}x+1=0\\x+5=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
b. \(2x^2+3x-5=0\)
⇔ 2x\(^2\) +5x - 2x -5 = 0
⇔ (2x\(^2\) - 2x)+ (5x - 5 ) = 0
⇔ 2x (x - 1) + 5 ( x - 1 ) = 0
⇔ ( 2x + 5 ) ( x - 1 ) = 0
⇔\(\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=\frac{-5}{2}\\x=1\end{matrix}\right.\)
\( a){x^2} + 6x + 5 = 0\\ \Leftrightarrow {x^2} + 5x + x + 5 = 0\\ \Leftrightarrow x\left( {x + 5} \right) + x + 5 = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 5 = 0\\ x + 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 5\\ x = 1 \end{array} \right.\\ b)2{x^2} + 3x - 5 = 0\\ \Leftrightarrow 2{x^2} + 5x - 2x - 5 = 0\\ \Leftrightarrow x\left( {2x + 5} \right) - \left( {2x + 5} \right) = 0\\ \Leftrightarrow \left( {2x + 5} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2x + 5 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{5}{2}\\ x = 1 \end{array} \right. \)
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